Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $k = \dfrac{-10r - 30}{-4r - 28} \div \dfrac{r + 1}{r^2 + 8r + 7} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $k = \dfrac{-10r - 30}{-4r - 28} \times \dfrac{r^2 + 8r + 7}{r + 1} $ First factor the quadratic. $k = \dfrac{-10r - 30}{-4r - 28} \times \dfrac{(r + 7)(r + 1)}{r + 1} $ Then factor out any other terms. $k = \dfrac{-10(r + 3)}{-4(r + 7)} \times \dfrac{(r + 7)(r + 1)}{r + 1} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ -10(r + 3) \times (r + 7)(r + 1) } { -4(r + 7) \times (r + 1) } $ $k = \dfrac{ -10(r + 3)(r + 7)(r + 1)}{ -4(r + 7)(r + 1)} $ Notice that $(r + 1)$ and $(r + 7)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ -10(r + 3)\cancel{(r + 7)}(r + 1)}{ -4\cancel{(r + 7)}(r + 1)} $ We are dividing by $r + 7$ , so $r + 7 \neq 0$ Therefore, $r \neq -7$ $k = \dfrac{ -10(r + 3)\cancel{(r + 7)}\cancel{(r + 1)}}{ -4\cancel{(r + 7)}\cancel{(r + 1)}} $ We are dividing by $r + 1$ , so $r + 1 \neq 0$ Therefore, $r \neq -1$ $k = \dfrac{-10(r + 3)}{-4} $ $k = \dfrac{5(r + 3)}{2} ; \space r \neq -7 ; \space r \neq -1 $